Geometric Series Sum Calculator
Sum a geometric sequence: enter the first term, common ratio and number of terms for the finite sum, nth term and infinite sum when it converges.
How it works
nth term aₙ = a₁ · r^(n−1) · Finite sum Sₙ = a₁ · (rⁿ − 1) ÷ (r − 1), r ≠ 1 · Infinite sum = a₁ ÷ (1 − r) when |r| < 1
A geometric series multiplies by a fixed ratio each step: 3, 6, 12, 24, … Its sum has a compact closed form, a₁(rⁿ − 1) ÷ (r − 1), which you can derive by subtracting r·Sₙ from Sₙ so that all the middle terms cancel. When the ratio has magnitude below 1 the terms shrink toward zero fast enough that even infinitely many of them add to a finite total, a₁ ÷ (1 − r); otherwise the sum grows without bound.
Worked example
For a first term of 3, ratio 2 and 8 terms: the last term is 3 × 2⁷ = 384, and the sum is 3 × (2⁸ − 1) ÷ (2 − 1) = 3 × 255 = 765. Because the ratio is 2, which is at least 1 in magnitude, the infinite series diverges — the terms keep doubling and never settle.
Sources: Geometric series — Wikipedia
Frequently asked questions
When does an infinite geometric series converge?
Only when the common ratio satisfies |r| < 1. Then the terms shrink geometrically and the total approaches a₁ ÷ (1 − r). If |r| ≥ 1 the terms do not shrink to zero and the sum grows without limit.
What happens when the ratio is exactly 1?
Every term equals the first term, so the series is just a₁ added n times: the sum is a₁ × n. The closed-form fraction is skipped because it would divide by zero.
Why does the finite-sum formula divide by r − 1?
It comes from the telescoping trick: Sₙ − r·Sₙ leaves only a₁ − a₁rⁿ, so Sₙ(1 − r) = a₁(1 − rⁿ). Dividing by (1 − r), or equivalently (r − 1) with signs flipped, isolates the sum.
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This calculator is for educational purposes. Double-check important results before acting on them.